Binary compounds are formed by ............... ............... elements.

What is the mass percent of potassium sulfate in solution if 78g of potassium sulfate is dissolved in 500 mL of water? (The dens

Debora [2.8K]

Answer:

13.5 %

Explanation:

First we<u> calculate the mass of 500 mL of water</u>, using <em>its density</em>:

Volume * Density = Mass

500 mL * 1.00 g/mL = 500 g

Then we <u>calculate the mass percent of potassium sulfate</u>, using the formula:

Mass of Potassium Sulfate / Total Mass * 100%

78 g / (78 + 500) g * 100 % = 13.5 %

6 0

2 years ago

A pharmaceutical company wants to test the efficiency of its new drug production techniques so they run 3 shifts of production f

spayn [35]

The percentage yield of the new production technique is 82.8%

<h3>What is the percentage yield?</h3>

Production is the procedure by which finished products are obtained form the raw materials. The production process involves the passing of raw materials through a certain procedure that involves the use of certain machines and equipment to give us the required products.

We are told in the question that there are three shifts;

Shift 1 produces 4562 grams

Shift 2 produces 5783 grams

Shift 3 produces 5247 grams

Average production from the three shifts = 4562 grams + 5783 grams + 5247 grams/3 = 5197 grams

The theoretical average yield is = 7000 grams + 7000 grams + 7000 grams/3 = 7000 grams

Now the percentage yield = actual yield/ theoretical yield * 100/1

percentage yield = 5197 grams/7000 grams * 100/1

percentage yield = 82.8%

Learn more about percentage yield :brainly.com/question/27492865

#SPJ1

3 0

2 years ago

How are energy and mass connected during the formation of an atom?

Jlenok [28]

I think the right answer for this question is option A. The energy absorbed so the mass will be increased.

6 0

3 years ago

Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --

frosja888 [35]

Answer :

(a) The anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.